Integrand size = 24, antiderivative size = 47 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=-\frac {\arctan \left (\frac {1-3 e^{2 x}}{\sqrt {2}}\right )}{6 \sqrt {2}}+\frac {1}{12} \log \left (1-2 e^{2 x}+3 e^{4 x}\right ) \] Output:
1/12*ln(1-2*exp(2*x)+3*exp(4*x))-1/12*arctan(1/2*(1-3*exp(2*x))*2^(1/2))*2 ^(1/2)
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=\frac {1}{12} \left (-\sqrt {2} \arctan \left (\frac {1-3 e^{2 x}}{\sqrt {2}}\right )+\log \left (1-2 e^{2 x}+3 e^{4 x}\right )\right ) \] Input:
Integrate[E^(4*x)/(1 - 2*E^(2*x) + 3*E^(4*x)),x]
Output:
(-(Sqrt[2]*ArcTan[(1 - 3*E^(2*x))/Sqrt[2]]) + Log[1 - 2*E^(2*x) + 3*E^(4*x )])/12
Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2720, 1142, 27, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{4 x}}{-2 e^{2 x}+3 e^{4 x}+1} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} \int \frac {e^{2 x}}{1-2 e^{2 x}+3 e^{4 x}}de^{2 x}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {1}{1-2 e^{2 x}+3 e^{4 x}}de^{2 x}+\frac {1}{6} \int -\frac {2 \left (1-3 e^{2 x}\right )}{1-2 e^{2 x}+3 e^{4 x}}de^{2 x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {1}{1-2 e^{2 x}+3 e^{4 x}}de^{2 x}-\frac {1}{3} \int \frac {1-3 e^{2 x}}{1-2 e^{2 x}+3 e^{4 x}}de^{2 x}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \int \frac {1}{-8-e^{4 x}}d\left (-2+6 e^{2 x}\right )-\frac {1}{3} \int \frac {1-3 e^{2 x}}{1-2 e^{2 x}+3 e^{4 x}}de^{2 x}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {6 e^{2 x}-2}{2 \sqrt {2}}\right )}{3 \sqrt {2}}-\frac {1}{3} \int \frac {1-3 e^{2 x}}{1-2 e^{2 x}+3 e^{4 x}}de^{2 x}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {6 e^{2 x}-2}{2 \sqrt {2}}\right )}{3 \sqrt {2}}+\frac {1}{6} \log \left (-2 e^{2 x}+3 e^{4 x}+1\right )\right )\) |
Input:
Int[E^(4*x)/(1 - 2*E^(2*x) + 3*E^(4*x)),x]
Output:
(ArcTan[(-2 + 6*E^(2*x))/(2*Sqrt[2])]/(3*Sqrt[2]) + Log[1 - 2*E^(2*x) + 3* E^(4*x)]/6)/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {\ln \left (1-2 \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{4 x}\right )}{12}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (6 \,{\mathrm e}^{2 x}-2\right ) \sqrt {2}}{4}\right )}{12}\) | \(38\) |
risch | \(\frac {\ln \left ({\mathrm e}^{2 x}-\frac {1}{3}+\frac {i \sqrt {2}}{3}\right )}{12}+\frac {i \ln \left ({\mathrm e}^{2 x}-\frac {1}{3}+\frac {i \sqrt {2}}{3}\right ) \sqrt {2}}{24}+\frac {\ln \left ({\mathrm e}^{2 x}-\frac {1}{3}-\frac {i \sqrt {2}}{3}\right )}{12}-\frac {i \ln \left ({\mathrm e}^{2 x}-\frac {1}{3}-\frac {i \sqrt {2}}{3}\right ) \sqrt {2}}{24}\) | \(70\) |
Input:
int(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x,method=_RETURNVERBOSE)
Output:
1/12*ln(1-2*exp(x)^2+3*exp(x)^4)+1/12*2^(1/2)*arctan(1/4*(6*exp(x)^2-2)*2^ (1/2))
Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {3}{2} \, \sqrt {2} e^{\left (2 \, x\right )} - \frac {1}{2} \, \sqrt {2}\right ) + \frac {1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \] Input:
integrate(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x, algorithm="fricas")
Output:
1/12*sqrt(2)*arctan(3/2*sqrt(2)*e^(2*x) - 1/2*sqrt(2)) + 1/12*log(3*e^(4*x ) - 2*e^(2*x) + 1)
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.47 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=\operatorname {RootSum} {\left (96 z^{2} - 16 z + 1, \left ( i \mapsto i \log {\left (8 i + e^{2 x} - 1 \right )} \right )\right )} \] Input:
integrate(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x)
Output:
RootSum(96*_z**2 - 16*_z + 1, Lambda(_i, _i*log(8*_i + exp(2*x) - 1)))
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, e^{\left (2 \, x\right )} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \] Input:
integrate(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x, algorithm="maxima")
Output:
1/12*sqrt(2)*arctan(1/2*sqrt(2)*(3*e^(2*x) - 1)) + 1/12*log(3*e^(4*x) - 2* e^(2*x) + 1)
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, e^{\left (2 \, x\right )} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \] Input:
integrate(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x, algorithm="giac")
Output:
1/12*sqrt(2)*arctan(1/2*sqrt(2)*(3*e^(2*x) - 1)) + 1/12*log(3*e^(4*x) - 2* e^(2*x) + 1)
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=\frac {\ln \left (3\,{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1\right )}{12}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}}{2}-\frac {3\,\sqrt {2}\,{\mathrm {e}}^{2\,x}}{2}\right )}{12} \] Input:
int(exp(4*x)/(3*exp(4*x) - 2*exp(2*x) + 1),x)
Output:
log(3*exp(4*x) - 2*exp(2*x) + 1)/12 - (2^(1/2)*atan(2^(1/2)/2 - (3*2^(1/2) *exp(2*x))/2))/12
Time = 0.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.81 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx=-\frac {\sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}+1}\, \sqrt {2}-2 e^{x} \sqrt {3}}{\sqrt {\sqrt {3}-1}\, \sqrt {2}}\right )}{12}-\frac {\sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}+1}\, \sqrt {2}+2 e^{x} \sqrt {3}}{\sqrt {\sqrt {3}-1}\, \sqrt {2}}\right )}{12}+\frac {\mathrm {log}\left (-e^{x} \sqrt {\sqrt {3}+1}\, \sqrt {2}+e^{2 x} \sqrt {3}+1\right )}{12}+\frac {\mathrm {log}\left (e^{x} \sqrt {\sqrt {3}+1}\, \sqrt {2}+e^{2 x} \sqrt {3}+1\right )}{12} \] Input:
int(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x)
Output:
( - sqrt(sqrt(3) + 1)*sqrt(sqrt(3) - 1)*atan((sqrt(sqrt(3) + 1)*sqrt(2) - 2*e**x*sqrt(3))/(sqrt(sqrt(3) - 1)*sqrt(2))) - sqrt(sqrt(3) + 1)*sqrt(sqrt (3) - 1)*atan((sqrt(sqrt(3) + 1)*sqrt(2) + 2*e**x*sqrt(3))/(sqrt(sqrt(3) - 1)*sqrt(2))) + log( - e**x*sqrt(sqrt(3) + 1)*sqrt(2) + e**(2*x)*sqrt(3) + 1) + log(e**x*sqrt(sqrt(3) + 1)*sqrt(2) + e**(2*x)*sqrt(3) + 1))/12